Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 8)
8.
The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:
Answer: Option
Explanation:
| P |  |
1 + | 20 |  |
n | > 2P |  |
|
6 | |
n | > 2. |
| 100 | 5 |
| Now, | |
6 | x | 6 | x | 6 | x | 6 | |
> 2. |
| 5 | 5 | 5 | 5 |
So, n = 4 years.
Discussion:
26 comments Page 1 of 3.
Narayana Rao Gartam said:
1 decade ago
Dear Jahnavi,
Amount = P(1+R/100)^n.
Sum of money is more than the double of Amount.
So, 2P < P(1+20/100)^n , Since R= 20%
P and P cancelled each other. So,
2< (1+20/100)^n.
2< (120/100)^n.
2<(6/5)^n.
Try n=1, 2, 3, 4, 5.
What least number can satisfy the 2<(6/5)^n.
Lets n=1 ( 2<1.2 so, wrong).
Lets n=2 ( 2<1.44 so, wrong).
Lets n=3 ( 2<1.728 so, wrong).
Lets n=4 ( 2<2.1036 so, right).
Answer is 4 years.
Amount = P(1+R/100)^n.
Sum of money is more than the double of Amount.
So, 2P < P(1+20/100)^n , Since R= 20%
P and P cancelled each other. So,
2< (1+20/100)^n.
2< (120/100)^n.
2<(6/5)^n.
Try n=1, 2, 3, 4, 5.
What least number can satisfy the 2<(6/5)^n.
Lets n=1 ( 2<1.2 so, wrong).
Lets n=2 ( 2<1.44 so, wrong).
Lets n=3 ( 2<1.728 so, wrong).
Lets n=4 ( 2<2.1036 so, right).
Answer is 4 years.
(1)
Kapil dev said:
8 years ago
Let principal be P.
P(1+R/100)^T>2P
P(1+20/100)^T>2P
(1+20/100)^T>2
(120/100)^T>2
1.2^T>2.
Now let's find out the minimum value of T for which the above equation becomes true.
If T = 1, 1.2T = 1.21 = 1.2
If T = 2, 1.2T = 1.22 * 1.4
If T = 3, 1.2T = 1.23 * 1.7
If T = 4, 1.2T = 1.24 * 2.07 which is greater than 2
Hence T = 4
i.e., the required number of years = 4.
P(1+R/100)^T>2P
P(1+20/100)^T>2P
(1+20/100)^T>2
(120/100)^T>2
1.2^T>2.
Now let's find out the minimum value of T for which the above equation becomes true.
If T = 1, 1.2T = 1.21 = 1.2
If T = 2, 1.2T = 1.22 * 1.4
If T = 3, 1.2T = 1.23 * 1.7
If T = 4, 1.2T = 1.24 * 2.07 which is greater than 2
Hence T = 4
i.e., the required number of years = 4.
(8)
Sohi said:
1 decade ago
Let x=(6/5)^n
(6/5)^n >2....................(1)
if we put value of n=1 then x=1.2.........condtion 1 is false.
if we put value of n=2 then x=1.44.........condtion 1 is still false.
if we put value of n=3 then x=1.728.........condtion 1 is false.
if we put value of n=4 then x=2.07.........now condtion 1 is true.
Hence answer is 4.
Thanx.
(6/5)^n >2....................(1)
if we put value of n=1 then x=1.2.........condtion 1 is false.
if we put value of n=2 then x=1.44.........condtion 1 is still false.
if we put value of n=3 then x=1.728.........condtion 1 is false.
if we put value of n=4 then x=2.07.........now condtion 1 is true.
Hence answer is 4.
Thanx.
Harini said:
2 years ago
Here they said that the money they put out in compound interest is greater than double of the money.
Let the money they put out in CI at a 20% rate of interest be P.
P(1+(20/100)^n > 2P,
(1+(20/100))^n > P,
(1.2)^n > 2.
taking log on both sides;
log(1.2)^n >log 2.
n> log 2/log 1.2.
n>3.8.
n = 4 yrs.
Let the money they put out in CI at a 20% rate of interest be P.
P(1+(20/100)^n > 2P,
(1+(20/100))^n > P,
(1.2)^n > 2.
taking log on both sides;
log(1.2)^n >log 2.
n> log 2/log 1.2.
n>3.8.
n = 4 yrs.
(16)
Ritesh said:
10 years ago
Hello @Anjali,
First take Amount = Rs. 5000 and Rate = 2%, Years = 1.
Then calculate SI, SI = PNR/100.
So, SI = 5000*1*2/100 = RS. 100.
[Here you are getting only interest].
Now calculate CI, CI = P(1+r)^n.
So, CI = 5000*(1+2/100)^1 = RS. 5100.
[Here we are getting Interest+Principal amount].
First take Amount = Rs. 5000 and Rate = 2%, Years = 1.
Then calculate SI, SI = PNR/100.
So, SI = 5000*1*2/100 = RS. 100.
[Here you are getting only interest].
Now calculate CI, CI = P(1+r)^n.
So, CI = 5000*(1+2/100)^1 = RS. 5100.
[Here we are getting Interest+Principal amount].
Pankaj said:
10 years ago
Very simple.
Suppose amount is Rs. 100.
Rate of interest is 12% annual.
Interest after 1st year-100*1.20 = 120.
Interest after 2nd year-120*1.20 = 144.
Interest after 3rd year-144*1.20 = 172.8.
Interest after 4th year-172.8*1.20 = 207.36.
So means to say it will take 4 years to double money.
Suppose amount is Rs. 100.
Rate of interest is 12% annual.
Interest after 1st year-100*1.20 = 120.
Interest after 2nd year-120*1.20 = 144.
Interest after 3rd year-144*1.20 = 172.8.
Interest after 4th year-172.8*1.20 = 207.36.
So means to say it will take 4 years to double money.
(7)
Lakshmanan sp said:
1 decade ago
AS p(6/5)^n>2p.
Next step (6/5)^n>2.
Next step ,take n = 1, then 6/5 = 1.2 therefore (1.2)^1 = 1.2.
|||rly for n = 2(1.2)^2 = 1.44.
For n = 3(1.2)^3 = 1.44*1.2 = 1.7288.
For n = 4(1.2)^4 = 1.7288*1.2 = 2.07.
Hence minimum no.of years is 4.
Next step (6/5)^n>2.
Next step ,take n = 1, then 6/5 = 1.2 therefore (1.2)^1 = 1.2.
|||rly for n = 2(1.2)^2 = 1.44.
For n = 3(1.2)^3 = 1.44*1.2 = 1.7288.
For n = 4(1.2)^4 = 1.7288*1.2 = 2.07.
Hence minimum no.of years is 4.
Sanjay Rajnedra Awate said:
7 years ago
For finding the time period in which a sum of money will double itself at R% rate of compound interest compounded annually, we generally use either of the following two formulas:
Time, T = 72 /R Years.
Time, T = 72 /R Years.
(4)
Shubham said:
9 years ago
What is the least number of complete years in which sum will become more than double itself at 12% per annum?
Given me the answer with clear explanation.
Given me the answer with clear explanation.
(1)
Muhammad Zunnoorain said:
9 years ago
We can also solve it as;
2P = P(1.2)^n.
2 = 1.2^n.
Applying lag on both side.
Log(2) = n*log(1.2).
n = log(2)/log(1.2).
n = 3.8 to nearest year 4.
2P = P(1.2)^n.
2 = 1.2^n.
Applying lag on both side.
Log(2) = n*log(1.2).
n = log(2)/log(1.2).
n = 3.8 to nearest year 4.
(2)
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